Why does the engine wind out when in B mode?

It would just change the speed at which the HV ECU spins MG1 to do the starting.

The engine ECU knows the crank speed of the engine, and the HV ECU knows the speeds and rotor positions of both MG1 and MG2 to eleven-bit precision (better than a tenth of a degree). It's perfectly capable of calculating, hmm, I want to crank the engine at 1,200 rpm, MG2 is currently doing x rpm, so I'm going to spin MG1 at y.

The tooth counts give you the torque relationships (through a formula no more complicated than the one you've given for motors). For calculating the power transmitted mechanically, you need both the rpm and the torque, so that's how they have to do with anything.

Edit: noticed this:

You've probably looked at motor/generator formulas more recently than I have, so I can't say for sure if this one is appropriate for the Prius MGs ... remember, they don't have a wound armature, or a stator supplying a DC field. The Prius armature has permanent magnets; the stator has three-phase AC windings.

If you're strongly interested in the details of the MGs, that might affect the formulas you're using; at the same time, as fuzzy1 suggests, it's probably a bit of a tangent to just understanding how the PSD functions.

 

When trying to understand how the ICE / MG! / MG2 / PSD system parts interplay, you can disregard everything inn this snippet except RPM and Power (not V).

Vs are adjusted or transformed between devices by SMPS (switch mode power supplies), so matching between devices is not an issue here.

Voltage and power are not equivalent.

Power = voltage * current = RPM * torque.

For this PSD, a planetary gear device, tooth count means everything.

This planetary gear device is a 3-shaft, or 3-terminal device, but it has only 2 degrees of freedom. Once you know the RPMs of any two shafts, simple algebra fixes and tells you the RPM of the remaining shaft. Torque is even simpler, once you know the torque on any shaft, simple algebra from those tooth ratios tells you the torque on both other shafts.

 

Back when I took physics, there was algebra involved. Have they changed that? Is f = ma now being taught as "well, when you push something harder, it kinda goes faster faster, unless it's like bigger"?

You seem to know enough electrical concepts to know that when power is flowing in electrical form, it is E ✕ I. The two are related in a way you explained rather nicely yourself back in #81 (81!) :

Exactly. That is, E is an electromotive force trying to make some current flow, but if no current actually flows, no power has moved, no matter how big E is. If current does flow, then the rate of that flow (I) times E gives you the power (dang, there's that algebra again).

Now what if you're talking about power not in electrical form, but the form of mechanical rotation?

Torque is a force trying to make something rotate. You can try as hard as you want, but if the thing doesn't rotate, no power has been transmitted, no matter how big the torque is. If the thing does rotate, then the rate of that rotation (rpm) times the torque gives you the power.

Any of that sound a little familiar?

You have 100 foot pounds of torque on your end of the lever. You have a torque of the same magnitude, but opposite sign, on the other end of the lever, so you have a balanced see-saw.

You usually can't get whole revolutions out of a see-saw because one end hits the ground (and the grinning kid on the other end gets launched). So for a better example, let's say you have a drum with a two-foot radius. There's a bunch of rope wound around it, and a 50 lb weight hanging off the rope.

Now, you can grab a four foot handle, attach it to the center of the drum, and start turning. You need your 25 lbs of force on the handle to just balance the hanging weight. By applying any tiny amount more torque, you can begin winding up the rope and lifting the weight.

After one revolution, you have lifted the weight by the circumference of the drum (2π times the 2 foot radius), about 13 feet. That was work! About 628 foot pounds of work (50 pounds raised 12.57 feet).

If you're able to keep that up and do 60 of those revolutions in a minute, why, you're doing 628 foot pounds of work per second—which would be 851 watts, or more than a horsepower. You probably can't keep that up; I don't know your build, but 10 rpm might be more realistic for that setup.

It's a little funny that both torque and work have the same dimensions (a force and a length) though they are different concepts. It's the same in SI: a newton-meter is a unit of torque, but a newton-meter is also a joule, the SI unit of work.

If you have a weight pulling by 1 N hanging from a drum with radius 1 m, and you turn that drum one revolution (with 1 Nm of torque), you've raised the weight 2π meters, and done 2π joules of work. If you can spin that drum 60 rpm (which you easily can), you're doing 2π joules of work every second, or 2π watts, since 1 W = 1 J/s.

So that's why power in rotary form is the torque times the rotation rate, just as in electrical form it is E times I. Trying to talk about rotary power without knowing the torque is just like trying to talk about electrical power without knowing the volts.

 

Quite right, as far as it goes; exactly as current does not indicate voltage in any way. In each case, you need to know both (rpm and torque, or current and voltage) to calculate power, which is the product of the two.

H.P. is a unit of power, not of torque. 1 H.P. is 550 ft-lb/sec as you provided yourself, or equally about 746 watts. Torque you could express as foot-pounds or newton-meters (or sometimes m-kgf, ugh) but never in HP or watts.

Combine a torque with a speed (in rpm), and then you can express it in HP or watts, because then it's power you're talking about.

• Power (in watts) = torque (in newton-meters) ✕ speed (in rpm) ✕2π ÷ 60
• Power (in HP) = torque (in foot-pounds) ✕ speed (in rpm) ✕2π ÷ 60 ÷ 550
• etc.

At various points in this thread you've been getting warmer, and right now you're burning hot.

Yes, "time when to apply more energy to MG2" (and to MG1, which also plays an important role) ...

That would be why each MG contains three sets of electromagnet coils, and why the inverter contains six transistor switches for each MG (a pair for each winding phase, wired as an H-bridge), and why each motor has such a sensitive resolver built in, reporting to the computer its exact speed and position to a 4096th of a revolution, so the computer knows exactly where each rotor permanent magnet is at each exact instant the computer wants to switch a pulse of current through any of the coils.

And the current that is set up by the computer through those switches, when passing through those coiled wires ...

... creates a magnetic field, which ...

... pulls on the permanent magnets in the rotor (the computer knows how hard it will pull, because it knows right where the magnets are in relation to the coils at that instant, thanks to the resolver) ...

... that pull is a force ... which you can multiply by the rotor radius, to get ...

... (wait for it) ...

... a torque!

In fact, you can plug in Techstream and ask the HV ECU how much torque it is commanding from MG1 and MG2 at any instant, and it'll tell you. It can do the math. "... time when to apply more energy ..." is exactly how it's done.

So yes, exactly, that's why the car has an HV ECU, not just a big red toggle switch.

The physics being covered in this thread really isn't concerned with torque ratings or specifications or maximums. Yes, the components involved have ratings and maximums and the engineers know what those are.

But when the gears of the PSD are doing what they do, splitting the torque at thing A into 28% at thing B and 72% at thing C, that has nothing to do with maximum figures in some engineer's spreadsheet somewhere, but only with the instantaneous torques, whatever they are, being experienced by thing A, thing B, and thing C at that moment. Gears live strictly in the now, and they can't read spreadsheets.

 

I think it this might be because this was a fairly shallow grade, averaging just 2.3%. At highway speeds, this is almost a neutral glide, not a strong charging opportunity. At 65-70 mph, it is just a glide without charging.

If this had been a much steeper grade, such as 5% (2600 feet in under 10 miles), or even 7% (same drop in just 7 miles), it would have made a far better charging test.

 

And totally irrelevant to this thread, because the SMPSs / inverters connecting all the high voltage parts (MG1, MG2, battery) can step voltage or current up or down at will to produce the desired power flow between these parts.

Please go back and review your basic engineering mechanics.

The algebra expresses how the engineering and physics of this device works. You can't throw it away.

 

Both of these are false. The units don't match up. Please go back and review your basic engineering mechanics.

No, it applies when they are spinning too.

 

When stepping voltage up or down, power remains nearly constant. 'Nearly' because there is a small bit of conversion loss to be subtracted out.

Don't forget to include the (implied, and intuitively obvious to the engineers and physicists in the crowd) RPM to radians/second conversion factor.

Algebra is symbolic math, not just numbers.

 

All of the above, depending on conditions.

The PSD is a 3-shaft device. Power can be split just about any way you can think of. And is, in different circumstances.

Sorry, I don't treat it as a 2-shaft in-line device. I prefer to stick with 3 shafts. 3 terminals. ICE, Battery, Wheels.

And except for proper ICE valve operation, spin directions don't particularly matter to this view, there are enough gear opportunities to switch things to fit packaging constraints. Just designate a normal forward and reverse at your convenience, so that the algebraic signs are consistent.